where $-t = r^{-3} (1 + 2r cos (3 \theta))$. Note that for $r = 1$, $t = 0$ when $\theta = \frac{2 \pi}{9}$ and $C(t) = \sqrt{3}$. But for the case $r = 1$ (which might not be interesting since we have used the determinant to renormalise the matrix) this solution set makes sense only provided $cos 3 \theta < 0.5$ and then

$C(t) = 2 cos (\frac{cos^{-1} (0.5 t)}{3})$

Observe that the solution condition states that $\theta > \frac{\pi}{9}$. Now observe that $\frac{2}{9} < \frac{\pi}{9}$ so this method is not directly helpful in analysing the lepton type matrix with $r = 1$. On the other hand, $\frac{1}{2} > \frac{\pi}{9}$ so the neutrino type cubic has three real solutions for $X$, but only $X_1 = 1.5644$ is positive. For a general positive real determinant $D$, $t = D + 2 cos (3 \theta))$ and the solution condition says that $cos (3 \theta) < \frac{2 - D}{2}$ which is less restrictive if $D$ is small.